# Thread: Integral convergence proof.

1. ## Integral convergence proof.

Prove that the integral $\displaystyle \int^{\infty}_3\frac{sinx\cdot lnx}{x^2\sqrt{x^2-4}}dx$ is converges.

Thanks!

2. Originally Posted by Also sprach Zarathustra
Prove that the integral $\displaystyle \int^{\infty}_3\frac{sinx\cdot lnx}{x^2\sqrt{x^2-4}}dx$ is converges.

Thanks!
$\displaystyle \left|\int^{t}_3\frac{\sin x\cdot \ln x}{x^2\sqrt{x^2-4}}dx\right|<\int^{t}_3\frac{\ln x}{x^2\sqrt{x^2-4}}dx$

Now $\displaystyle \ln x<x$ and $\displaystyle \frac{1}{\sqrt{x^2-4}}<\frac{1}{\sqrt{x}} \; \forall \; x\geq3$

Thus we get $\displaystyle \left|\int^{t}_3\frac{\sin x\cdot \ln x}{x^2\sqrt{x^2-4}}dx\right|<\int^{t}_3 \frac{x}{x^2\cdot\sqrt{x}}dx=\int^{t}_3\frac{1}{x^ {3/2}}dx = \frac{2}{\sqrt{3}}-\frac{2}{\sqrt{t}}$

So finally $\displaystyle \lim_{t\to\infty}\left|\int^{t}_3\frac{\sin x\cdot \ln x}{x^2\sqrt{x^2-4}}dx\right|<\lim_{t\to\infty}\left(\frac{2}{\sqrt {3}}-\frac{2}{\sqrt{t}}\right)=\frac{2}{\sqrt{3}}$

3. Originally Posted by Also sprach Zarathustra
Prove that the integral $\displaystyle \int^{\infty}_3\frac{sinx\cdot lnx}{x^2\sqrt{x^2-4}}dx$ is converges.

Thanks!
Maybe...

$\displaystyle \int^{\infty}_3|\frac{sinx\cdot lnx}{x^2\sqrt{x^2-4}}dx|<\int^{\infty}_3\frac{1\cdot (x)}{x^2\sqrt{x^2-4}}dx<\int^{\infty}_3\frac{1}{x\sqrt{x}}dx<\int\fr ac{1}{x^{1.5}}dx$

and so on...

4. I late!